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lord [1]
3 years ago
13

The enthalpy change for the explosion of ammonium nitrate with fuel oil is –7198 kJ for every 3 moles of NH4NO3. What is the ent

halpy change for 1.0 mole of NH4NO3 in this reaction?
Chemistry
1 answer:
anastassius [24]3 years ago
7 0

Answer:

−2399.33 kJ

Explanation:

If NH₄NO₃ reacts with fuel oil to give a ΔH of -7198 for every 3 moles of NH₄NO₃

What is the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction

∴ For every 1 mole, we will have \frac{1}{3} of the total enthaply of the 3 moles

so, to determine the 1 mole; we have:

\frac{1}{3}*(-7198kJ)

= −2399.33 kJ

∴ the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction = −2399.33 kJ

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In an experiment, 34.8243g of copper (II) nitrate hydrate, Cu(NO3)2•zH2O was heated to a constant mass of 27.0351g. Calculate th
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Answer:

1) The mass of water lost = 7.7892 grams

2) Z = 3: Cu(NO3)2*3H2O

Explanation:

<u>Step 1: </u>Data given

Mass of copper (II) nitrate hydrate, Cu(NO3)2•zH2O = 34.8243 grams

Mass of substance after heating = 27.0351 grams

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of H2O = 18.02 g/mol

<u>Step 2:</u> Calculate mass of water

The mass of water is the mass lost after heating.

Mass water = 34.8243 - 27.0351 = 7.7892 grams of water

<u>Step 3:</u> Calculate moles of Cu(NO3)2

Moles Cu(NO3)2 = Mass Cu(NO3)2 / Molar mass Cu(NO3)2

Moles Cu(NO3)2 = 27.0351 grams / 187.56 g/mol

Moles Cu(NO3)2 = 0.144 moles

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Z = 3

This means we have 3 water molecules in the formula. This makes the formula ofthe hydrate: Cu(NO3)2*3H2O

5 0
3 years ago
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